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3 years ago

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

late: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

2 answers:

krek1111 [17]3 years ago

7 0

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

$\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}$

$10*2 = \sqrt{1 + 8f^2 - 1$

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

$\frac{V_1}{\sqrt{g*y_1}} = 7.416$

$V_1 = 7.416 *\sqrt{32.2*1}$

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

$V_2 = \frac{3666.66}{80*10}$

= 4.208 ft/sec

Froude number, F2 =

$\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}$

F2 = 0.235

d) $El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}$

$El = \frac{(10-1)^3}{4*1*10}$

$= \frac{9^3}{40}$

= 18.225ft

e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

Savatey [412]3 years ago

6 0

Answer:

(a). 42.1 ft/s, (b). 3366.66 ft^3/s, (c). 0.235, (d). 18.2 ft, (e). 3.8 ft.

Explanation:

The following parameters are given in the question above and they are;

Induced hydraulic jump, j = 80 ft wide channel, and the water depths on either side of the jump are 1 ft and 10 ft. Let k1 and k2 represent each side of the jump respectively.

(a). The velocity of the faster moving flow can be calculated using the formula below;

k1/k2 = 1/2 [ √ (1 + 8g1^2) - 1 ].

Substituting the values into the equation above a s solving it, we have;

g1 = 7.416.

Hence, g1 = V1/ √(L × k1).

Therefore, making V1 the subject of the formula, we have;

V1 = 7.416× √ ( 32.2 × 1).

V1 = 42.1 ft/s.

(b). R = V1 × j × k1.

R = 42.1 × 80 × 1.

R = 3366.66 ft^3/s.

(c). Recall that R = V2 × A.

Where A = 80 × 10.

Therefore, V2 = 3366.66/ 80 × 10.

V2 = 4.21 ft/s.

Hence,

g2 = V1/ √(L × k2).

g2 = 4.21/ √ (32.2 × 10).

g2 = 0.235.

(d). (k2 - k1)^3/ 4 × k1k2.

= (10 - 1)^3/ 4 × 1 × 10.

= 18.2 ft.

(e).The critical depth;

[ (3366.66/80)^2 / 32.2]^ 1/3.

The The critical depth = 3.80 ft.

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